Quadratic Functions

[sparklemetink]

Here is the question... Find a quadratic function f(x)=ax^2 +bx+c whose x-intercepts are -4 and 2 and whose Range is [-18, 0].

My thought is to work backwards to see if I could come up with the quadratic function. All I have come up with so far is this...

Knowing I have x-intercepts of -4 and 2. I came up with (x+4)(x-2)
And knowing I have a Range of [-18, 0], I came up with sqrt x(x+18)

Not sure how to put these together or if I'm going in the right direction. Any help would be appreciated. Thank you.

 

[mmm4444bot]

Find a quadratic function f(x) = ax^2 + bx + c whose … Range is [-18, 0].

The graph of any quadratic function is a parabola. Parabolas continue forever, so the Range of all quadratic functions is infinite.

The only way to constrain the Range of a quadratic function to [-18, 0] is to restrict the Domain.

Are you sure that the requested Range is [-18, 0], in this exercise ?

If the Range were to be [-18, ?), then your exercise would make more sense, to me.

Otherwise, you'll have to restrict the Domain to [-4, 2].

In either event, one way to find the values of the parameters a, b, and c is to set up and solve a system of three equations.

Subsitute the following values into the generic quadratic equation for x and y, and you'll end up with a system of three equations containing a, b, and c.

x = -4, y = 0

x = 2, y = 0

x = -1, y = -18

These last values come from the vertex coordinates. Since you know that the x-intercepts are -4 and 2, the x-coordinate of the vertex must be halfway between them. That's -1.

 

[mmm4444bot]

I was just fixing lunch, when I thought of another approach, for finding the parameter values.

From the given x-intercepts, we have:

y = a(x + 4)(x - 2)

You know the vertex coordinates are (-1, 18).

Substitute those values for x and y, and solve for a.

We also know that the x-coordinate of the vertex is -b/(2a).

Therefore, b must equal 2a, in this exercise, because -b/(2a) = -1.

Now that you have the values for both a and b, finding the value of c is trivial.

Please show your work, if you would like more help with this exercise.

And, please confirm that the Range is constrained to [-18, 0] .

Cheers ~ Mark

 

[sparklemetink]

Thank you Mark. I think your second approach is more with what my teacher is looking for. Yes, the Range is [-18, 0]. I will work on this and get back to you if I have any questions.

Thank you.

 

[mmm4444bot]

Yes, the Range is [-18, 0].

Thank you.

So, in addition to reporting the polynomial definition of function f (with your values of a, b, and c), you must also report the restricted Domain; otherwise, your answer will be incomplete because the parameters a, b, and c themselves do nothing to restrict f(x) from taking on positive values. Only a specific Domain will do that.

The image below shows the part of the parabola with domain [-5,3], and you can image what the effect would be on the graph, if the domain were to be further restricted to [-4,2].

Oh, and contrary to what my second post implies, once you find the value of a, you can simply multiply out a(x + 4)(x - 2), to get the polynomial. There's no need to first find b and c.

Cheers !

[attachment=0:30yowhpg]constrained.JPG[/attachment:30yowhpg]